This comment from Ada gave me the idea for the following solution. Unfortunately, as Ada correctly pointed out later on, the approach is flawed as the angular bisectors of the triangles I mention are not those of the quadrilateral.
The below anser is therefore incorrect and only here for reference.
The three internal angle bisectors of a triangle meet in a point inside the triangle. In a similar fashion, two external and one internal angle bisector will meet in a point outside the triangle.
So consider triangle $ABC$. Take external bisectors for $\angle A$ and $\angle C$, and the internal bisector for $\angle B$. These three lines meet in a point called $Y$. Using $D$ instead of $B$ you can teel that the internal bisector fro $\angle D$ goes through $Y$ as well.
Now you can reverse your argumentation: you know that the point $Y$ is defined by the fact that the two internal bisectors for $\angle B$ and $\angle D$ intersect in this point. So now you can see that the external angle bisectors at $\angle A$ and the line $AY$ have to be one and the same. Likewise for $C$ instead of $A$. So now you know these connecting lines to be external angle bisectors, which of course are orthogonal to the corresponding internal angle bisectors.